1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Dual vector space - Lagrange Interpolating Polynomial

  1. Feb 21, 2013 #1
    I think I solved it a week ago, but I didn't write down all the things and I want to be sure of doing the things right, plus the excersise of writing it here in latex helps me a loot (I wrote about 3 threads and didn't submited it because writing it here clarified me enough to find the answer for my self xD). So feel free to correct me.

    Lets be [itex]\alpha _{0},...,\alpha _{n} \in K, \alpha _{i} \neq \alpha _{j}[/itex] if [itex] i\neq j[/itex]

    To every [itex]i[/itex], [itex]0 \leq i \leq n[/itex], we define [itex]\varepsilon _{\alpha _i}:K_{n}[X]\rightarrow K [/itex] as [itex]\varepsilon _{\alpha _i}(P)=P(\alpha _i)[/itex].

    i) Prove that [itex]B = \left \{ \epsilon _{\alpha _{0}},...,\epsilon _{\alpha _{n}} \right \}[/itex] is a basis of [itex](K_{n}[X])^{*}[/itex]

    ii) Lets be [itex]B = \left \{ P_{0},...,P_{n} \right \}[/itex] the basis of [itex]K_{n}[X][/itex] such that [itex]B^{*}=B_{1}[/itex]. Prove that the polynomial

    [itex]P = \sum _{i=0}^{n} \beta _i . P_i[/itex]

    is the only polynomial in K[X] of grade minor or equal to n such that,

    [itex]\forall i, 0 \leq i \leq n, P(\alpha _{i}) = \beta _{i}[/itex]

    This polynomial is named the Lagrange's Interpolating Polynomial (irrelevant :P).

    iii) Prove that there're real numbers [itex]\left \a_{0},...,a_{n}{ \right \}[/itex] such that, to all [itex]P \in \mathbb{R}_{n}[X][/itex]
    [itex]\int _{0}^{1}P(x)dx=\sum _{i=0}^{n}a_{i}.P(\alpha _{i})[/itex]
    Find [itex]a_{0}, a_{1}[/itex] and [itex]a_{2}[/itex] when [itex]n = 2, a_{0}=1, a_{1}=\frac{1}{2}[/itex] and [itex] a_{2}=0[/itex]

    Relevant proposition:
    Lets be V a K-vector space of dimensión n, and lets be V* it's dual space.
    Lets be [itex]B_{1}=\left \{ \varphi _{1},...\varphi _{n} \right \}[/itex] a basis of V*. Then there's only one basis [itex]B = \left \{ v _{1},...v _{n} \right \}[/itex] of V such that:

    i) Suppose that:
    [itex]\sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}=0[/itex]
    Then, to all [itex]P \in K_n[X][/itex], we have: [itex]\sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}(P)=\sum _{i=0}^{n}P(\alpha _{i})=0[/itex]
    To every [itex]\forall i, 0 \leq i \leq n[/itex]
    We take the polynomial:
    [itex]\prod _{i=0,{i\neq }j}^{n}(X-\alpha _{i})[/itex]
    And evaluating it we know [itex]\alpha _{j}=0[/itex], because [itex]P(\alpha _{i})=0[/itex] to every i≠j [itex](P(\alpha _{j})[/itex] can't be zero because [itex]\alpha _{j} \neq \alpha _{i}[/itex] if i≠j
    Then, we have the linear independency of the set, and as n+1 elements, then it's generated space has dimention n+1 and it's a subset of [itex](K_{n}[X])*[/itex], so it's a basis of [itex](K_{n}[X])*[/itex] (because it also has dimension n+1).
    ii) So we know [itex]B_{1}[/itex] is a basis of [itex]K_n[X][/itex], then for the proposition we know that there's only one basis
    [itex]B= \left \{ P_{0},...,P_{n} \right \}[/itex] such that [itex]B^{*}=B_{1}[/itex]
    So, by the definition of dual basis, [itex]\epsilon _{\alpha _{i}}(P_{j})=P_{j}(\epsilon _{\alpha _{0}})=[/itex]
    1 if i=j
    0 if i≠j
    Then [itex]P(\epsilon _{\alpha _{j}})=\sum _{i=0}^{n} \beta _i . P_i(\epsilon _{\alpha _{j}})=\beta _{i}[/itex]

    I think I'm stuck in 3, but I'll think later after sleeping, I'm probably failing due to being so sleepy.
    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 22, 2013 #2
    I didn't find the edit button in my first post, I assume someone blocked it so I bump the thread when I finish it instead of buring it in the bottom of the forum's abyss.
    If [itex]P=\sum _{i=0}^{n}\beta _{i} P_{i} [/itex] then
    [itex]\int _{0}^{1}\sum _{i=0}^{n} \beta _{i} . P_{i} (x)dx = \sum _{i=0}^{n} \beta _{i}.\int _{0}^{1}P_{i} (x)dx[/itex]
    [itex]\sum _{i=0}^{n}a_i.P(\alpha _{i})=\sum _{i=0}^{n}a_i.\sum _{j=0}^{n} \beta _{i} P_{i} (\alpha _{i})=[/itex]
    [itex]\sum _{i=0}^{n}a_{i}.\beta _{i}[/itex]
    Let's be [itex]a_{i}=\int _{0}^{1}P_{i} (x)dx[/itex] to all [itex]0 \leq i \leq n[/itex]
    [itex]\sum _{i=0}^{n} \frac{\beta _{i}}{i+1}=\sum _{i=0}^{n}a_{i}.\beta _{i}[/itex]

    Let's do find [itex]a_{0}, a_{1}[/itex] and [itex]a_{2}[/itex] when [itex]n = 2, a_{0}=1, a_{1}=\frac{1}{2}[/itex] and [itex] a_{2}=0[/itex]
    P_{0}(\alpha _{0})=1\\
    P_{0}(\alpha _{1})=0\\
    P_{0}(\alpha _{2})=0
    As we know two roots of [itex]P_{0}[/itex] we only need to multiply it for a constant to make it equal to 1 when evaluated in 1.



    So we only need to evaluate the integrals of these polynomials.

    PD: The excersise is from http://mate.dm.uba.ar/~jeronimo/algebra_lineal/AlgebraLineal.pdf page 115 Ejercise 16, this isn't the book used in my Linear Algebra's course in the dual vector space subject but I find the ejercises were pretty interesting to do.
    Last edited: Feb 22, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook